undecidability of the domino problem · e. jeandel and p. vanier, undecidability of the domino...

Undecidability of the Domino Problem

E. Jeandel and P. Vanier

Loria (Nancy), LACL (Créteil), France

November 20-24

E. Jeandel and P. Vanier, Undecidability of the Domino Problem 1/86

The theorem (again)

Theorem (Berger 1964 (PhD), 1966 (Memoirs of the AMS))There is no algorithm that decides, given a tileset τ , if τ tiles the plane.

Last time

Algorithms can be represented by Turing machinesTuring machines can be encoded into tilings

However, the encoding has a default: it needs a “catalyst” tile to launchthe computation.

Proof of the Undecidability of the Domino Problem

Proofs fall into 4 categories:Berger (64)-Robinson (71) constructionsAanderaa-Lewis (74)Kari (07)Durand-Romashchenko-Shen (08)

Many, many variants of Berger-Robinson.

Two different ways to solve the caveat we saw on tuesday:

Force the catalyst tile to appear(Berger,Robinson,Aanderaa-Lewis,DRS)Use a different encoding of Turing machines (Kari)

We will present here the two methods

Remark

Everything I’m presenting now is very well knownThe lecture notes have a third proof (Aanderaa-Lewis) which isalmost completely unknown

If you know everything about Kari and Robinson, read the lecturenotes!

1 Berger-Robinson

2 The Kari constructionEncoding of TM

3 Conclusion

Historical Notes

Inspired by the proof of Berger and Robinson, using ideas ofOllinger for simplificationRobinson’s proof is not that different from Berger

Berger has an aperiodic set of 103 tiles in his PhD

The Catalyst tile

We want to force a specific tile to appear so that it can kickstart thecomputation

Reminder

A tileset τ tiles the plane iff it tiles arbitrarily large squares

CorollaryIf every tiling of the plane contains a tile t , then there exists n s.t. tappears in every n × n square

The Catalyst tile

The specific tile t should appear everywhere.

There should be infinitely many computations in one tiling of the plane.

How to do it

To obtain the Undecidability of the Domino Problem:

Find an aperiodic tileset τLook at occurrences of a specific tile t inside this τBuild computations around this tile, hoping that they do not overlap

How to do it

How do you obtain an aperiodic tileset ?

Robinson-Berger: Use substitution-like tilings.

Substitutions

DefinitionA substitution is a map φ from A to An×n for some n.Given a substitution φ and an initial pattern w , we can iterate φ on w toobtain arbitrarily large patterns.

Example

Substitutive tilesets

DefinitionThe subshift Sφ associated to φ is the set of colorings of the plane thatlook like φk (w) for some large value of k .

(Of course not an exact definition)

DefinitionA tileset τ is substitutive if tilings by τ look like Sφ up to recoloring.

A lot of different substitutive tilesets in the literature. But how can youobtain them ?

Intrinsically substitutive tileset

DefinitionA tileset τ is intrinsically substitutive if the substitution φ is defineddirectly on the tiles of τ , and encoded directly into the local constraintsof τ .

What do we mean exactly ?φ(w) is a valid tiling iff w is a valid tiling

There is a one-to-one correspondence between the colors on the eastside of τ and colors on the east side of φ(τ).

We can desubstitute: For every tiling x by τ , there exists y s.tφ(y) = x

All? tilings of a 2× 2 square are of the form φ(t) for t ∈ τ

Theorem (Durand-Levin-Shen 2004, Ollinger 2008)There is an intrinsically substitutive tileset

These are the first proven examples. However Berger’s set of 103 tilesIS intrinsically substitutive, and Robinson’s set ISN’T.

The tileset of Ollinger

A tiling

How does it work

First layer

The first layer forces tiles to be grouped into 2× 2 squares.

Second layer

(16 different tiles)

(4 different tiles)

The second layer is the image of the first layer by the substitution(more on this later)

The third layer

The third layer codes the “chair substitution”.

Substitution

φ is a 2× 2 substitution. When substituting, the original tile is “found”on the bottom-left of the 2× 2 square.

Layer 1 of w is encoded into layer 2 of φ(w)

Layer 2 and 3 of w are enlarged into φ(w)

Layer 0→ Layer 1

Every φ(w) is of the form:

Layer 1→ Layer 2

Layer 1 is enlarged into Layer 2

7→ 7→

Layer 2→ Layer 2

Layer 2 is propagated into the adjacent tiles

7→ 7→

Layer 3→ Layer 3

Layer 3 is also propagated

7→ 7→

Two examples

Theorem

Theorem (Ollinger 2008)The tileset τ is intrinsically substitutive

Proof: Just? inspect all 2× 2 patterns.

Theorem

The fact the tileset is substitutive proves it is aperiodic but does notexplain everything

Why do we need the third layer ?

What are the rules between the layers ?

The substitution again

Every φ(w) is of the form:

7→ A H

A: any tileC: corner tileH: horizontal tile: propagate signal from A horizontallyV : vertical tile: propagate signal from A vertically

What if we have something of the form

The corner at the top right of the square determines what kind of tileswe can put at the bottom left of the square

That’s what the third layer is ensuring.

More details

7→ A H

What color appear on the blue part determines which corner is in C.

Rules for V

Here are the possible rules for the tile at the top left (tile of type V ):

Similar rules for the tile at the bottom right.

Theorem

Theorem (Ollinger 2008)The tileset τ is instrinsically substitutive, and we more or lessunderstand why.

What do we do now ?

We will encode computations inside the tilings s.t. there is a tiling iffsome Turing Machine does not halt.

First, we simplify the tilings by looking only at the red color at layer 2

What we have

We obtain squares in which we could put a Turing machine

Squares have a designated corner in which we could put the catalytictile.

Problem

Squares intersect

Solution: Keep only half the squares.Technically: two different shades of red (dark/light), dark red can onlycross light red.

Problem

Squares inside squares.

We cannot change this

Technical trick due to Robinson: Use on bigger square only the spacethat is not hindered by the smaller squares.

What to do now

We now put our Turing machine in the space left blankIt will be initialized by the bottom left corner of a square, which isalways free (blank)The free (blank) space inside a square is not connected, howeverwe can use the gray cells to transmit information(colors) betweentwo blank cells on the same row/columnLot of technical details.

Suppose there is a tiling of the planeThen there are tilings of arbitrarily large red squaresThese squares contains computation of the Turing machine for anarbitrarily long timeTherefore the Turing machine does not halt

Suppose the TM does not haltBuild a tiling of the plane by putting in each red square thebeginning of the computation of the TM.

Therefore there exists a tiling iff the TM does not halt.

Remarks

Essentially Robinson’s proofRobinson is not substitutive. Ollinger is a cover of Robinson that isintrinsically substitutive.

Berger’s proof builds infinite vertical strips rather than squaresThe vertical strips overlap, so same trick is used

Berger’s set of 104 103 tiles is almost the same asRobinson/Ollinger !

1 Berger-Robinson

3 Conclusion

Kari construction

Main idea :

Use a different encoding of Turing machines into tilings that do nothave the problems we encountered

Problems (from last time)

Erratic configurations:

tilings with no head per rowtilings with more than one head per rowtilings may no start from the initial statetilings using only the blank tile

New encoding

This new encoding (to be defined) forces that there is exactly one headper row.

Which problem remains ?

The Immortality Problem

Theorem (Hooper 1966)There is no algorithm that decides, given a Turing machine, if thereexists SOME configuration on which it runs forever.

Configuration can be arbitrarily (arbitrarily initial state, arbitrarily initialfilling of the tape)

New encoding + Hooper Theorem = WIN

Note: Proof of Hooper’s Theorem is very hard.

Kari’s idea

Two steps:

It is very easy to simulate affine maps with tilesets

A Turing Machine can be encoded into piecewise affine maps.

How to encode Turing machines

A Turing Machine can be encoded into piecewise affine maps.

Idea: use the moving tape model.

Suppose the states and symbols of the Turing machine are integers.

1 2 0 1 2 2 0 1 0 3 2

Suppose the states and symbols of the Turing machine are integers.

1 2 0 1 2 2 0 1 0 3 2

Code it into:L = 0.021 . . .R = 0.12201032 . . .S = 1

Encoding

Every configuration of the Turing machine will be encoded into a tuple(L,R,S) ∈ R3, representing the left part, the right part, and the state ofthe configuration.

How does it work

The TM is in state q and reads an x is the same as

S ∈ [q,q + 1[ and R ∈ [x/10, x/10 + 0.1[

How does it work

We replace the x by a y is the same as

R ← R + (y − x)/10

How does it work

We move the head to the right is the same as

L← (L + x)/10

R ← 10(R − x/10)

How does it work

If we are in state q and read x , we output y , change to state q′ and wemove to the right:

f (L,R,S) = ((L + y)/10,10(R − x/10),q′)

if S ∈ [q,q + 1[ and R ∈ [x/10, x/10 + 0.1[

Result

A Turing machine is essentially the same as a piecewise affine map

Caveats

The base in which we represent the tape should be strictly biggerthan the alphabet.

To avoid 0.099999 · · · = 0.100000 . . .

Some inputs of the piecewise affine map do not correspond to“real” tapes

This is usually not a problem

Result

Theorem (Hooper 1966)There is no algorithm that decides, given a Turing machine, if thereexists SOME configuration on which it runs forever.

TheoremThere is no algorithm that decides, given a 3D piecewise affine map f ,if there exists x s.t f n(x) is defined for all n.

Now it remains to..

Kari’s idea

It is very easy to simulate affine maps with tilesets

(if they have rational coefficients)

We will only explain how to simulate a map f : x → ax + b. Addinglinear constraints (x < c) or going to higher dimensions is easy.

One row of colors will represent a real number x .

One row of a tiling represents transitions between two rows of color(the colors at the south, and the colors at the north)

We will design the tileset so that one row of the tiling will represent theoperation x → f (x).

Question

How do we represent real numbers as biinfinite sequences ?

The β-sequence

DefinitionThe β-sequence for a number x is the biinfinite sequence

βn(x) = bnxc − b(n − 1)xc

Examples:β(0) = . . . 000000000000000000000000 . . . .β(1/2) = . . . 010101010101010101010101 . . . .β(1/3) = . . . 100100100100100100100100 . . . .β(√

2− 1) = . . . 001010010100101010010100 . . . .

Properties of β-sequences

βn(x) ∈ {bxc, bx + 1c}n consecutive bits of β(x) gives a good approximation of x :∑k+n

i=k+1 βi(x)n

=b(k + n)xc − bkxc

n∈ [x − 1

n, x +

Backwards

We have obtained a map from real numbers to biinfinite sequences of0 and 1.

We need a reverse map, for example:

r(w) = lim infn→+∞

∑ni=−n wi

2n + 1

Notice that r(β(x)) = x .

Main theorem

TheoremFor every q ∈ Q there exists a tileset τ s.t.

For every x, there exists a tiling of a row with β(x) at the bottomand β(qx) at the topIn a tiling of a row by τ , we have r(t) = qr(b) where t and bdenote the colors on the top and bottom of the row.

First item: The tileset simulates multiplication by q if fed the right input.Second item: It also simulates multiplication by q on bad inputs.

Idea of the proof

The tileset τ consists of all tiles of the form

c c′b

with c, c′ ∈ [−q,1] andqa + c = b + c′

Ideally we would like qa = b. c represents a carry from the past, and c′

a carry we will give to our future.

First point

This works if we have β(x) on bottom and β(qx) on top

qb(n − 1)xc − bq(n − 1)xc qbnxc − bqnxc

βn(qx)

βn(x)

Second point

Suppose we have t on top and b at bottom.Then

i=−n

bi + c−n =n∑

i=−n

ti + cn

Thereforeq∑n

i=−n bi

2n + 1=

∑ni=−n ti

2n + 1+

cn − c−n

2n + 1And

qr(b) = r(t)

as cn − c−n = O(1)

Slight problem

Note: we need a finite tileset

We only need carries that can occur on the β-sequences. If q = n/mwe only need carries in [−q,1] that are rational fractions of m, andthere are finitely many of them.

Main theorem

TheoremFor every piecewise affine map f with rational coefficients, there existsa tileset τ s.t.

For every x, there exists a tiling of a row with β(x) at the bottomand β(f (x)) at the topIn a tiling of a row by τ , we have r(t) = f (r(b)) where t and bdenote the colors on the top and bottom of the row.

End of the proof

Let f be a piecewise affine map.

Suppose there exist n s.t. f n(x) is defined for all n.

Then there exists a tiling of a half plane with β(f n(x)) as the bottomcolor of the n-th row

(By the first point)

Therefore there exists a tiling of the plane by compactness

End of the proof

Let f be a piecewise affine map.

Suppose there exist a tiling of the plane

Let bn be the color at the bottom of the n-th row.

Then r(bn) = f (r(bn−1)) and therefore r(bn) = f n(x) for x = r(b0)

(By the second point)

Therefore f n(x) is defined for all n.

End of the proof

There is a tiling of the plane by τ iff there exists x s.t. f n(x) is definedfor all n.

No algorithm can solve the latter problem, therefore no algorithm cansolve the former.

Final remark

The proof is considerably simpler, but we have to admit the theorem ofHooper.

Somehow, the difficulty has been shifted into the proof of thetheorem of HooperThe proof of Hooper’s theorem uses tricks similar to Robinson (inparticular substitutions)

1 Berger-Robinson

3 Conclusion

Conclusion

Two different proofs of the Undecidability of the Domino Problem

One more proof in the lecture notesOne last proof will be given in the full version of the lecture notes.

undecidability of the domino problem · e. jeandel and p. vanier, undecidability of the domino...

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